Integrand size = 42, antiderivative size = 156 \[ \int \frac {A+B x+C x^2}{\sqrt {a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3}} \, dx=-\frac {2 \left (A b^2-a (b B-a C)\right ) (a+b x)}{b^3 \sqrt {a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3}}+\frac {2 (b B-2 a C) (a+b x)^2}{b^3 \sqrt {a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3}}+\frac {2 C (a+b x)^3}{3 b^3 \sqrt {a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3}} \] Output:
-2*(A*b^2-a*(B*b-C*a))*(b*x+a)/b^3/(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^(1/ 2)+2*(B*b-2*C*a)*(b*x+a)^2/b^3/(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^(1/2)+2 /3*C*(b*x+a)^3/b^3/(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^(1/2)
Time = 0.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.38 \[ \int \frac {A+B x+C x^2}{\sqrt {a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3}} \, dx=\frac {2 (a+b x) \left (-3 A b^2-8 a^2 C+a b (6 B-4 C x)+b^2 x (3 B+C x)\right )}{3 b^3 \sqrt {(a+b x)^3}} \] Input:
Integrate[(A + B*x + C*x^2)/Sqrt[a^3 + 3*a^2*b*x + 3*a*b^2*x^2 + b^3*x^3], x]
Output:
(2*(a + b*x)*(-3*A*b^2 - 8*a^2*C + a*b*(6*B - 4*C*x) + b^2*x*(3*B + C*x))) /(3*b^3*Sqrt[(a + b*x)^3])
Time = 0.43 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.59, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2008, 1140, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {A+B x+C x^2}{\sqrt {a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3}} \, dx\) |
\(\Big \downarrow \) 2008 |
\(\displaystyle \frac {(a+b x)^{3/2} \int \frac {C x^2+B x+A}{(a+b x)^{3/2}}dx}{\sqrt {(a+b x)^3}}\) |
\(\Big \downarrow \) 1140 |
\(\displaystyle \frac {(a+b x)^{3/2} \int \left (\frac {\sqrt {a+b x} C}{b^2}+\frac {b B-2 a C}{b^2 \sqrt {a+b x}}+\frac {A b^2-a (b B-a C)}{b^2 (a+b x)^{3/2}}\right )dx}{\sqrt {(a+b x)^3}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(a+b x)^{3/2} \left (-\frac {2 \left (A b^2-a (b B-a C)\right )}{b^3 \sqrt {a+b x}}+\frac {2 \sqrt {a+b x} (b B-2 a C)}{b^3}+\frac {2 C (a+b x)^{3/2}}{3 b^3}\right )}{\sqrt {(a+b x)^3}}\) |
Input:
Int[(A + B*x + C*x^2)/Sqrt[a^3 + 3*a^2*b*x + 3*a*b^2*x^2 + b^3*x^3],x]
Output:
((a + b*x)^(3/2)*((-2*(A*b^2 - a*(b*B - a*C)))/(b^3*Sqrt[a + b*x]) + (2*(b *B - 2*a*C)*Sqrt[a + b*x])/b^3 + (2*C*(a + b*x)^(3/2))/(3*b^3)))/Sqrt[(a + b*x)^3]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x _Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 0]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Simp[((a + b*x)^Exp on[Px, x])^p/(a + b*x)^(Expon[Px, x]*p) Int[u*(a + b*x)^(Expon[Px, x]*p), x], x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; !IntegerQ[p] && PolyQ[Px, x ] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Time = 0.38 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.46
method | result | size |
risch | \(\frac {2 \left (C x b +3 B b -5 C a \right ) \left (b x +a \right )^{2}}{3 b^{3} \sqrt {\left (b x +a \right )^{3}}}-\frac {2 \left (A \,b^{2}-a b B +C \,a^{2}\right ) \left (b x +a \right )}{b^{3} \sqrt {\left (b x +a \right )^{3}}}\) | \(71\) |
gosper | \(-\frac {2 \left (b x +a \right ) \left (-C \,b^{2} x^{2}-3 B \,b^{2} x +4 C a b x +3 A \,b^{2}-6 a b B +8 C \,a^{2}\right )}{3 b^{3} \sqrt {b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}}}\) | \(80\) |
default | \(-\frac {2 \left (b x +a \right ) \left (-C \,b^{2} x^{2}-3 B \,b^{2} x +4 C a b x +3 A \,b^{2}-6 a b B +8 C \,a^{2}\right )}{3 b^{3} \sqrt {b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}}}\) | \(80\) |
orering | \(-\frac {2 \left (b x +a \right ) \left (-C \,b^{2} x^{2}-3 B \,b^{2} x +4 C a b x +3 A \,b^{2}-6 a b B +8 C \,a^{2}\right )}{3 b^{3} \sqrt {b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}}}\) | \(80\) |
trager | \(-\frac {2 \left (-C \,b^{2} x^{2}-3 B \,b^{2} x +4 C a b x +3 A \,b^{2}-6 a b B +8 C \,a^{2}\right ) \sqrt {b^{3} x^{3}+3 a \,b^{2} x^{2}+3 b \,a^{2} x +a^{3}}}{3 \left (b x +a \right )^{2} b^{3}}\) | \(82\) |
Input:
int((C*x^2+B*x+A)/(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^(1/2),x,method=_RETU RNVERBOSE)
Output:
2/3*(C*b*x+3*B*b-5*C*a)*(b*x+a)^2/b^3/((b*x+a)^3)^(1/2)-2/b^3*(A*b^2-B*a*b +C*a^2)*(b*x+a)/((b*x+a)^3)^(1/2)
Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.62 \[ \int \frac {A+B x+C x^2}{\sqrt {a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3}} \, dx=\frac {2 \, \sqrt {b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}} {\left (C b^{2} x^{2} - 8 \, C a^{2} + 6 \, B a b - 3 \, A b^{2} - {\left (4 \, C a b - 3 \, B b^{2}\right )} x\right )}}{3 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \] Input:
integrate((C*x^2+B*x+A)/(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^(1/2),x, algor ithm="fricas")
Output:
2/3*sqrt(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*(C*b^2*x^2 - 8*C*a^2 + 6 *B*a*b - 3*A*b^2 - (4*C*a*b - 3*B*b^2)*x)/(b^5*x^2 + 2*a*b^4*x + a^2*b^3)
\[ \int \frac {A+B x+C x^2}{\sqrt {a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3}} \, dx=\int \frac {A + B x + C x^{2}}{\sqrt {\left (a + b x\right )^{3}}}\, dx \] Input:
integrate((C*x**2+B*x+A)/(b**3*x**3+3*a*b**2*x**2+3*a**2*b*x+a**3)**(1/2), x)
Output:
Integral((A + B*x + C*x**2)/sqrt((a + b*x)**3), x)
Time = 0.06 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.56 \[ \int \frac {A+B x+C x^2}{\sqrt {a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3}} \, dx=-\frac {2 \, A}{\sqrt {b x + a} b} + \frac {2 \, {\left (b^{2} x^{2} + 3 \, a b x + 2 \, a^{2}\right )} B}{{\left (b x + a\right )}^{\frac {3}{2}} b^{2}} + \frac {2 \, {\left (b^{3} x^{3} - 3 \, a b^{2} x^{2} - 12 \, a^{2} b x - 8 \, a^{3}\right )} C}{3 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3}} \] Input:
integrate((C*x^2+B*x+A)/(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^(1/2),x, algor ithm="maxima")
Output:
-2*A/(sqrt(b*x + a)*b) + 2*(b^2*x^2 + 3*a*b*x + 2*a^2)*B/((b*x + a)^(3/2)* b^2) + 2/3*(b^3*x^3 - 3*a*b^2*x^2 - 12*a^2*b*x - 8*a^3)*C/((b*x + a)^(3/2) *b^3)
Time = 0.13 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.58 \[ \int \frac {A+B x+C x^2}{\sqrt {a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3}} \, dx=-\frac {2 \, {\left (C a^{2} - B a b + A b^{2}\right )}}{\sqrt {b x + a} b^{3} \mathrm {sgn}\left (b x + a\right )} + \frac {2 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} C b^{6} - 6 \, \sqrt {b x + a} C a b^{6} + 3 \, \sqrt {b x + a} B b^{7}\right )}}{3 \, b^{9} \mathrm {sgn}\left (b x + a\right )} \] Input:
integrate((C*x^2+B*x+A)/(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^(1/2),x, algor ithm="giac")
Output:
-2*(C*a^2 - B*a*b + A*b^2)/(sqrt(b*x + a)*b^3*sgn(b*x + a)) + 2/3*((b*x + a)^(3/2)*C*b^6 - 6*sqrt(b*x + a)*C*a*b^6 + 3*sqrt(b*x + a)*B*b^7)/(b^9*sgn (b*x + a))
Time = 12.82 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.52 \[ \int \frac {A+B x+C x^2}{\sqrt {a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3}} \, dx=-\frac {2\,\sqrt {a^3+3\,a^2\,b\,x+3\,a\,b^2\,x^2+b^3\,x^3}\,\left (8\,C\,a^2+4\,C\,a\,b\,x-6\,B\,a\,b-C\,b^2\,x^2-3\,B\,b^2\,x+3\,A\,b^2\right )}{3\,b^3\,{\left (a+b\,x\right )}^2} \] Input:
int((A + B*x + C*x^2)/(a^3 + b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x)^(1/2),x)
Output:
-(2*(a^3 + b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x)^(1/2)*(3*A*b^2 + 8*C*a^2 - C *b^2*x^2 - 6*B*a*b - 3*B*b^2*x + 4*C*a*b*x))/(3*b^3*(a + b*x)^2)
Time = 0.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.29 \[ \int \frac {A+B x+C x^2}{\sqrt {a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3}} \, dx=\frac {\frac {2}{3} b^{2} c \,x^{2}-\frac {8}{3} a b c x +2 b^{3} x -\frac {16}{3} a^{2} c +2 a \,b^{2}}{\sqrt {b x +a}\, b^{3}} \] Input:
int((C*x^2+B*x+A)/(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)^(1/2),x)
Output:
(2*( - 8*a**2*c + 3*a*b**2 - 4*a*b*c*x + 3*b**3*x + b**2*c*x**2))/(3*sqrt( a + b*x)*b**3)